Python | pycurl sends wildcard as path
Im currently writing a program which connects to a cam using RTSP. The python script is using pycurl.
I managed to make OPTIONS and DESCRIBE work. However SETUP is not working.
> Choose request or command: SETUP * Found bundle for host 192.168.0.60: 0x7f8e70a0b650 [can pipeline] * Re-using existing connection! (#0) with host 192.168.0.60 * Connected to 192.168.0.60 (192.168.0.60) port 554 (#0) > SETUP * RTSP/1.0 CSeq: 3 Transport: RTP/AVP;unicast;client_port=38274-38275 User-Agent: LibVLC/3.0.8 (LIVE555 Streaming Media v2018.02.18) Accept: */* < RTSP/1.0 400 Bad Request < Allow: OPTIONS, DESCRIBE, SETUP, TEARDOWN, PLAY, PAUSE, GET_PARAMETER, SET_PARAMETER,USER_CMD_SET < * The CSeq of this request 3 did not match the response 0 * Connection #0 to host 192.168.0.60 left intact Traceback (most recent call last): File "/Volumes/DATA/htdocs/forensik/forensik_curl/venv/curl.py", line 209, in <module> rtsp.handle(cmd) File "/Volumes/DATA/htdocs/forensik/forensik_curl/venv/curl.py", line 99, in handle self.sendRequest(pycurl.RTSPREQ_SETUP, header) File "/Volumes/DATA/htdocs/forensik/forensik_curl/venv/curl.py", line 53, in sendRequest self.crl.perform() pycurl.error: (85, 'The CSeq of this request 3 did not match the response 0') * Closing connection 0
I think the problem is that there is a wildcard * in the request path
SETUP * RTSP/1.0
To set the URL is use
Any idea what I can do ?
Thanks in advance!